Calculus

Infinite Sequences and Series

Comparison Tests

The \(N\)th term test, generally speaking, does not guarantee convergence of a series. Convergence or divergence of a series is proved using sufficient conditions. The comparison tests we consider below are just the sufficient conditions of convergence or divergence of series.

The Comparison Tests

Let \(\sum\limits_{n = 1}^\infty {{a_n}} \) and \(\sum\limits_{n = 1}^\infty {{b_n}} \) be series such that \(0 \lt {a_n} \le {b_n}\) for all \(n.\) Then the following comparison tests hold:

The Limit Comparison Tests

Let \(\sum\limits_{n = 1}^\infty {{a_n}} \) and \(\sum\limits_{n = 1}^\infty {{b_n}}\) be series such that \({a_n}\) and \({b_n}\) are positive for all \(n.\) Then the following limit comparison tests are valid:

Example 2.

We use the comparison test. Note that \({\large\frac{{{n^2} – 1}}{{{n^4}}}\normalsize} \) \( {\large\frac{{{n^2}}}{{{n^4}}}\normalsize} \) \(= {\large\frac{1}{{{n^2}}}\normalsize}\) for all positive integers \(n.\) As \(\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^2}}}\normalsize} \) is a \(p\)-series with \(p = 2 \gt 1\), it converges. Hence, the given series also converges by the comparison test.